Question

INSTRUCTIONS: Read the questions carefully and answer the following items accordingly. Show your solution properly. Q7 A technician made an experiment of an AC circuit in an electrical lab. The source voltage is AC with a frequency of 60Hz. He collected the data in the form of a phasor diagram as shown Figure 3. This phasor diagram is an impedance phasor. The values in the diagram are reactances, a resistance and current. The current in this case is the reference phasor at 0 degrees. Answer the following: a) Based on your knowledge about the behavior of the R, L and C and its principles applied in AC circuits, in the form of essay interpret carefully the diagram and describe in detail about the given information in Figure 3. (6 marks) b) Draw and label the final phasor diagram involving the impedance phasor Z and the phase angle. (2 marks) c) Calculate: i) The impedance in polar form, ii) phase angle between the E and I, iii) RMS values of Vr. V., and Vc, and iv) and the instantaneous equations of the e and i. (4 marks) Phonor Diagram of Series RLC AC Circuit 80 ohms 10 A 30 ohms 35 ohms - 2

Answer 1

Qiz hiven, the of - GOH 2 current in the rose sence phasor at oo (as the Element are behaviour of the R,L&L iron liners (bassing though origin), Passive (stose of diggipate Energy , Bilateral / Portion of graph both side.). 3on IOA Die on 1 350 Fig. 3 T 11.14 II aluadount ( Bilateral passive) Passing through colgin. to Voltage case-I i Whey Case-I :- when DC applied 12, C&C R> It dissipate Energy in me foom of hoat L cloesinf allow sudden Change of current but in Steady state alt af short ckt. ca doesnt allow suddan change in huf in steady state act as open ckt. AG is applied to cat in Rilec. It difsipate Energy in the form of heat L lags current by goo but load & voltage by 90° currens by goo hut lagß voltage by to reads given fig. 3, X= (x: Anductive tre - 1350 (XL: capacitive XL = json reactance) greactance) R= 3 on 1 1 - 10 LO From the fig. current leads too by cabretitive neuetanle and lagh 90° by anductive reactanle.

the IR=302 161 the phapor dingram involving sore. Ampedance phapor (al & bhaka angie (D) x = IBO, x = -735 z = R+XL - Xcls - 30+ 454 In - $6.30° (x2-x2) R z- 4.08 £ 56.300 - ZLO. tany (x2-xe) o Les til the in Impedana pourr form. - RJ (x₂-xe) in - For Anductive loud. 2= R + j(XC-XL) in - for capacitive lood. Z z = (304 Jurin - An rectangular foon z = 54.08L56.50 Un polus foom (i.) Phase Angle blu Ed I - 108 in I.10 Lo since, nomas boom i el 1 reference) E = ZI 54.08 256.30 x 10 Loo 540.8 L 56.30° Angle bw Et forms it 5-6.30°. l'ail RMS IR L60 x RLoo SOLO x 30 Loo UR 300 Loo value of Romis value of XL X. ILOO since, series A.LL Ckt I 18 Same VL- 80 L90° X loloo - Xcx lolo a 35L-90 x 10 Loo - 350L-90v 800 Lgo. V of

VR gamis 300 Lov, NL = 800 L90o v Tim's Verms 350L-gov. Liv.) the E anal in tuntuneous enrundion of and I E = Em losto +40) & I - im Coswt tol Ima 2 Ir.mis = √2x10 - 14.14 A w = Qaf - 2RxGo - 376.gg a 377 rad /see Em - Eng.ms - Q x 54.08 = 76.48 ~ take radian. - 56:30° = 56.300 K -0.982 reid have to 180 E = 76.48 Cos(377t +0.982) v 14.14 COS( 3991 A Hence, the instantaneous woltage and instantane- са дост. аде calculated Dus E € 1 = 76.48 Cos(377t +0.982) 14.14 Cos1377t ) A sonce, voltage load morant by 0.982 rad.